Find the coordinates of the points on the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z+1}{2}$ which are at a distance of $\sqrt{11}$ units from the origin.

Find the coordinates of the points on the line (x)/(1) = (y-1)/(2) = (z+1)/(2) which are at a distance of √(11

Question

\frac{x}{1} = \frac{y-1}{2} = \frac{z+1}{2}

CBSECBSE Class XII Board 2023Subjective· 2mImportance★★★★★

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Concept understanding — Distance From Point To Line

Distance from a Point to a Line

The distance from a point to a line is the shortest distance — the length of the perpendicular dropped from the point onto the line. In 3D we compute it with vectors and the cross product.

Let the line be $\vec{r}=\vec{a}+\lambda\vec{b}$ (a point $A$ with position vector $\vec{a}$ , direction $\vec{b}$ ), and let $P$ be the given point with position vector $\vec{p}$ .

The idea

Look at the triangle formed by $A$ , $P$ and the foot of the perpendicular $M$ . The segment $\overrightarrow{AP}=\vec{p}-\vec{a}$ is the hypotenuse, and the perpendicular distance $d=PM$ is the side opposite the angle $\theta$ between $\overrightarrow{AP}$ and the line:

d = |\overrightarrow{AP}|\sin\theta.

But the cross product already contains $\sin\theta$ : $|\overrightarrow{AP}\times\vec{b}| = |\overrightarrow{AP}|\,|\vec{b}|\sin\theta$ . Dividing by $|\vec{b}|$ isolates the distance.

d = \frac{|(\vec{p}-\vec{a})\times\vec{b}|}{|\vec{b}|}

Example

Distance of $P(1,2,3)$ from the line $\vec{r}=(\hat{i}+\hat{j})+\lambda(2\hat{i}-\hat{j}+2\hat{k})$ .

Here $\vec{a}=(1,1,0)$ , $\vec{b}=(2,-1,2)$ , and $\overrightarrow{AP}=\vec{p}-\vec{a}=(0,1,3)$ .

$\overrightarrow{AP}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\0&1&3\\2&-1&2\end{vmatrix} = (2+3)\hat{i} - (0-6)\hat{j} + (0-2)\hat{k} = 5\hat{i}+6\hat{j}-2\hat{k}.$ …

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We parameterize the given line as $( \lambda, 2\lambda+1, 2\lambda-1 )$ and use the distance formula from the origin, $\sqrt{x^2+y^2+z^2} = \sqrt{11}$ , to solve for $\lambda$ . This yields $\lambda = \pm 1$ , giving the points $(1, 3, 1)$ and $(-1, -1, -3)$ .

The problem asks us to find specific points on a given line in 3D space that are a certain distance from the origin. The core idea is to represent any point on the line using a single variable (a parameter) and then use the distance condition to solve for that parameter.

A line in 3D can be given in symmetric form, like $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$ . This form tells us that the line passes through the point $(x_0, y_0, z_0)$ and is parallel to the direction vector $\vec{d} = \langle a, b, c \rangle$ . To work with individual points on this line, it's often easiest to convert it into parametric form. We do this by setting each ratio equal to a parameter, say $\lambda$ . This allows us to express $x, y, z$ coordinates of any point on the line as functions of $\lambda$ .

Once we have a general point $(x(\lambda), y(\lambda), z(\lambda))$ on the line, we can use the standard 3D distance formula. The distance $D$ between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$ . For the distance from the origin $(0,0,0)$ to a point $(x,y,z)$ , this simplifies to $D = \sqrt{x^2 + y^2 + z^2}$ . We will set this distance equal to the given value, $\sqrt{11}$ , and solve for $\lambda$ .

Parameterize the line. The given equation of the line is $\frac{x}{1} = \frac{y-1}{2} = \frac{z+1}{2}$ . To express any point on this line in terms of a single variable, we set each ratio equal to a parameter, say $\lambda$ :

\frac{x}{1} = \lambda \implies x = \lambda

\frac{y-1}{2} = \lambda \implies y-1 = 2\lambda \implies y = 2\lambda + 1

\frac{z+1}{2} = \lambda \implies z+1 = 2\lambda \implies z = 2\lambda - 1

So, any point $P$ on the line can be represented by its coordinates $(\lambda, 2\lambda+1, 2\lambda-1)$.

2. Apply the distance condition.

We are given that the points are at a distance of $\sqrt{11}$ units from the origin $(0,0,0)$ .

The distance $D$ from the origin to a point $(x,y,z)$ is given by the formula:

> [!FORMULA]

> $D = \sqrt{x^2 + y^2 + z^2}$

In our case, $D = \sqrt{11}$, so we have:

\sqrt{x^2 + y^2 + z^2} = \sqrt{11}

Squaring both sides to simplify the equation:

x^2 + y^2 + z^2 = 11

Substitute parametric coordinates and solve for $\lambda$ . Now, substitute the parametric expressions for $x, y, z$ from Step 1 into the distance equation from Step 2:

(\lambda)^2 + (2\lambda+1)^2 + (2\lambda-1)^2 = 11

Expand the squared terms: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 16 on this concept.

CBSE 20261 markMCQ
$Q. The length of perpendicular drawn from point (2, 5, 7) on line x 1 = y 0 = z 0 is 1 (A) 2 (B) 5 (C) 74 (D) 78$
›Reveal solutionSolution
The perpendicular distance from a point to a line in 3D is found by projecting the vector from a point on the line to the given point onto the direction vector, then using the Pythagorean theorem. For point $(2,5,7)$ and the line $\frac{x}{1} = \frac{y}{0} = \frac{z}{0}$ , the distance is $\sqrt{74}$ , so the correct option is (C).

Concept and Intuition

The distance from a point to a line in 3D is the length of the perpendicular segment connecting the point to the line. This is not the same as the distance along any slant path — it's the shortest possible distance.

Think of it this way: if you stand at a point in space and look at a line, the shortest path to reach that line is to walk straight towards it at a right angle. That perpendicular distance is what we calculate.

The key idea: take any point $A$ on the line, form the vector $\overrightarrow{AP}$ from $A$ to the given point $P$ , then project $\overrightarrow{AP}$ onto the direction vector $\vec{d}$ of the line. The component of $\overrightarrow{AP}$ perpendicular to $\vec{d}$ gives the perpendicular distance.
$P$
This works because the cross product magnitude gives the area of the parallelogram formed by $\overrightarrow{AP}$ and $\vec{d}$ , and dividing by $|\vec{d}|$ gives the height (perpendicular distance) of that parallelogram.

Step-by-Step Solution

1. Identify the line and a point on it.

The line is given as $\frac{x}{1} = \frac{y}{0} = \frac{z}{0}$ . This means:

Direction ratios are $(1, 0, 0)$ — the line runs along the x-axis.

The line passes through the origin $(0, 0, 0)$ because when $x=0$ , $y=0$ , $z=0$ satisfies the equation.

So we have:

Point on line: $A = (0, 0, 0)$

Direction vector: $\vec{d} = (1, 0, 0)$

Given point: $P = (2, 5, 7)$

2. Find the vector from $A$ to $P$ .
$\overrightarrow{AP} = P - A = (2-0, 5-0, 7-0) = (2, 5, 7)$
3. Compute the cross product $\overrightarrow{AP} \times \vec{d}$ .
$\overrightarrow{AP} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 5 & 7 \\ 1 & 0 & 0 \end{vmatrix}$
Expanding:

$\hat{i}$ component: $(5)(0) - (7)(0) = 0$

$\hat{j}$ component: $-( (2)(0) - (7)(1) ) = -(0 - 7) = 7$

$\hat{k}$ component: $(2)(0) - (5)(1) = -5$

So $\overrightarrow{AP} \times \vec{d} = (0, 7, -5)$

Tip

Notice that since $\vec{d} = (1,0,0)$ is along the x-axis, the cross product simply picks out the y and z components of $\overrightarrow{AP}$ with a sign swap. This is a shortcut: for a line along the x-axis, the perpendicular distance is just $\sqrt{y^2 + z^2}$ of the point relative to the line.

4. Find the magnitude of the cross product.

$|\overrightarrow{AP} \times \vec{d}| = \sqrt{0^2 + 7^2 + (-5)^2} = \sqrt{0 + 49 + 25} = \sqrt{74}$ …
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CBSE 2026Set 65/1/11 markMCQ
$(2, 5, 7)$
›Reveal solutionSolution
To find the perpendicular distance from a point to a line, we identify a general point on the line, form a vector from the given point to this general point, and use the condition that this vector must be perpendicular to the line's direction vector. This allows us to find the specific point on the line (the foot of the perpendicular) and then calculate the distance. The length of the perpendicular is $\boxed{\sqrt{74}}$ .

When we talk about the "length of the perpendicular from a point to a line," we are essentially looking for the shortest distance between that point and any point on the line. Imagine dropping a plumb line from the given point straight down to the line; the length of that plumb line is what we need to find.

The core idea is that the shortest distance occurs along a line segment that is perpendicular to the given line. If we can find the exact point on the line where this perpendicular meets it (often called the "foot of the perpendicular"), then calculating the distance between the two points becomes straightforward using the standard 3D distance formula.

Here's how we approach this:

Represent a general point on the line: Any point on the given line can be expressed using a single parameter.

Form a vector: Create a vector connecting the given point to this general point on the line.

Apply perpendicularity: The key insight is that this connecting vector must be perpendicular to the direction vector of the line. In 3D geometry, two vectors are perpendicular if and only if their dot product is zero. This condition will allow us to find the specific value of the parameter.

Find the foot of the perpendicular: Substitute the parameter value back into the general point's coordinates to get the coordinates of the foot of the perpendicular.

Calculate the distance: Use the distance formula between the given point and the foot of the perpendicular.

Let's apply this method to the given problem.

Identify the given point and the line's properties.

The given point is $P = (2, 5, 7)$ .

The equation of the line is $\dfrac{x}{1} = \dfrac{y}{0} = \dfrac{z}{0}$ .

This symmetric form tells us two crucial things:

A point on the line (when $x=0, y=0, z=0$ ) is $A = (0, 0, 0)$ .

The direction vector of the line, $\vec{d}$ , has components given by the denominators: $\vec{d} = (1, 0, 0)$ .

Note

The line $\frac{x}{1} = \frac{y}{0} = \frac{z}{0}$ is a special case. It represents the x-axis itself, as it passes through the origin $(0,0,0)$ and has a direction along the x-axis.

Represent a general point on the line.

Let $Q$ be any general point on the line. Using the parametric form of the line, $x = 0 + 1\lambda$ , $y = 0 + 0\lambda$ , $z = 0 + 0\lambda$ , where $\lambda$ is a scalar parameter.

So, a general point on the line is $Q = (\lambda, 0, 0)$ .

Form the vector connecting the given point to the general point on the line.

The vector $\vec{PQ}$ connects point $P(2, 5, 7)$ to point $Q(\lambda, 0, 0)$ .

$\vec{PQ} = Q - P = (\lambda - 2, 0 - 5, 0 - 7) = (\lambda - 2, -5, -7)$ .

Apply the perpendicularity condition to find $\lambda$ .

For $\vec{PQ}$ to be the perpendicular from $P$ to the line, $\vec{PQ}$ must be perpendicular to the direction vector of the line, $\vec{d} = (1, 0, 0)$ .

The dot product of two perpendicular vectors is zero.
$\vec{u} = (u_1, u_2, u_3)$
So, $\vec{PQ} \cdot \vec{d} = 0$ :

$(\lambda - 2)(1) + (-5)(0) + (-7)(0) = 0$

$\lambda - 2 + 0 + 0 = 0$

$\lambda - 2 = 0$

$\lambda = 2$ .

Find the coordinates of the foot of the perpendicular. …
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CBSE 2026Set ANNUAL1 markMCQ
$(3, -5)$
(a) $\dfrac{5}{3}$
(b) $\dfrac{3}{5}$
(c) 0
(d) None of these
›Reveal solutionSolution

Apply $d = \dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$ with $A=3, B=-4, C=-26$ and the point $(3,-5)$ .

Distance from a point $(x_1,y_1)$ to the line $Ax+By+C=0$ :

$d = \dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$

Here $A=3, B=-4, C=-26$ , and $(x_1,y_1)=(3,-5)$ : …

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CBSE 2025Set ANNUAL1 markMCQ
$(6, 7)$
(a) 5
(b) 7
(c) 9
(d) none of these
›Reveal solutionSolution

The perpendicular distance from $(6,7)$ to $3x+4y+9=0$ works out to $11$ , which does not match any of the numeric options given, so the answer is "none of these."

The distance from a point $(x_1,y_1)$ to a line $ax+by+c=0$ is $d = \dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$ .

Here $a=3$ , $b=4$ , $c=9$ , $(x_1,y_1)=(6,7)$ :

$d = \dfrac{|3(6)+4(7)+9|}{\sqrt{3^2+4^2}} = \dfrac{|18+28+9|}{\sqrt{9+16}} = \dfrac{55}{5} = 11$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$6x - 7y = 11$
(a) $11/85$
(b) $11\sqrt{85}$
(c) $11\sqrt{85}/85$
(d) $\sqrt{85}$
›Reveal solutionSolution

The distance of $6x-7y=11$ from the origin is $\dfrac{11\sqrt{85}}{85}$ .

Write the line as $6x - 7y - 11 = 0$ . The distance from the origin $(0,0)$ to $ax+by+c=0$ is $d = \dfrac{|c|}{\sqrt{a^2+b^2}}$ .

Here $a=6$ , $b=-7$ , $c=-11$ : $d = \dfrac{|-11|}{\sqrt{6^2+(-7)^2}} = \dfrac{11}{\sqrt{36+49}} = \dfrac{11}{\sqrt{85}}$ . …

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CBSE 2025Set ANNUAL1 markMCQ
$4x - 3y = 7$
(a) 1
(b) 5
(c) 12
(d) 19
›Reveal solutionSolution

The distance between $4x-3y=7$ and $4x-3y=12$ is $1$ .

For two parallel lines $ax+by=c_1$ and $ax+by=c_2$ (same $a,b$ ), the distance between them is $d = \dfrac{|c_2-c_1|}{\sqrt{a^2+b^2}}$ .

…

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CBSE 2023Set ANNUAL1 markMCQ
$ax+by+c=0$
(a) $\left|\dfrac{d-c}{\sqrt{a^2+b^2}}\right|$
(b) $\left|\dfrac{d-c}{\sqrt{a^2-b^2}}\right|$
(c) $\dfrac{d-c}{\sqrt{ab}}$
(d) $d-c$
›Reveal solutionSolution

The distance between two parallel lines with the same $a,b$ but different constants $c,d$ is $\left|\dfrac{d-c}{\sqrt{a^2+b^2}}\right|$ .

For two parallel lines $ax+by+c=0$ and $ax+by+d=0$ (same slope, since $a,b$ match), the perpendicular distance between them is derived by picking any point on one line and applying the point-to-line distance formula to t …

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CBSE 2023Set ANNUAL1 markMCQ
$p$
(a) $x\cos\alpha - y\sin\alpha = p$
(b) $x\sin\alpha + y\cos\alpha = p$
(c) $x\cos\alpha + y\sin\alpha = p$
(d) None of these
›Reveal solutionSolution

The line is $x\cos\alpha+y\sin\alpha=p$ ; option (c).

If the perpendicular from the origin has length $p$ and makes angle $\alpha$ with the $x$ -axis, the foot of the perpendicular is $(p\cos\alpha,\ p\sin\alpha)$ and the line perpendicular to that direction is the no …

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CBSE 2022Set ANNUAL1 mark
$Q. The length of perpendicular from (0, 0) to 3x + 4y + 15 = 0 is ............ .$
›Reveal solutionSolution

Applying the point-to-line distance formula with $(0,0)$ gives $3$ .

Distance from $(x_1, y_1)$ to line $ax+by+c=0$ :

$d = \frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$

Here $(x_1,y_1) = (0,0)$ , and the line is $3x+4y+15=0$ ( $a=3, b=4, c=15$ ): …

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CBSE 2022Set ANNUAL1 markMCQ
$Q. What would be the distance of the origin from the line BC?$
(a) 1 km
(b) 1.2 km
(c) 1.4 km
(d) 2 km
›Reveal solutionSolution

Apply the point-to-line distance formula to the origin and the line $x-y+2=0$ found in the previous part.

The line BC is $x-y+2=0$ (from part iii). Distance of a point $(x_0,y_0)$ from line $Ax+By+C=0$ is $\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$ .

Here $A=1,B=-1,C=2$ , and the point is the origin $(0,0)$ :

…

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CBSE 2022Set ANNUAL1 markMCQ
$(2, -3, -1)$
(a) $2$
(b) $3$
(c) $6$
(d) $7$
›Reveal solutionSolution

Apply the point-to-plane distance formula to get $\frac{14}{7}=2$ .

Distance of $(x_1,y_1,z_1)$ from $ax+by+cz+d=0$ is $\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}$ .

…

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CBSE 2022Set ANNUAL1 markMCQ
$2x - 3y + 4z = 6$
(a) $\frac{6}{\sqrt{35}}$
(b) $\frac{6}{\sqrt{37}}$
(c) $\frac{6}{\sqrt{29}}$
(d) $\frac{6}{\sqrt{31}}$
›Reveal solutionSolution

Use $\frac{|d|}{\sqrt{a^2+b^2+c^2}}$ for $ax+by+cz=d$ .

Write the plane as $2x-3y+4z-6=0$ . Distance from $(0,0,0)$ :

…

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Q.Find the coordinates of the points on the line x1=y−12=z+12\frac{x}{1} = \frac{y-1}{2} = \frac{z+1}{2}1x​=2y−1​=2z+1​ which are at a distance of 11\sqrt{11}11​ units from the origin.

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Q.Find the coordinates of the points on the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z+1}{2}$ which are at a distance of $\sqrt{11}$ units from the origin.

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